This value of z substituted in (11) gives y=9. And these values of z and y, substituted in (6), give Equating (4) with (5); and (4) with (6), we have Equations (7) and (8) when reduced become Having found, we readily find y and z to be y=60; z=120. ELIMINATION BY SUBSTITUTION. (85.) There is still another method of elimination. 1. Suppose we have given the two equations Substituting this value of y in (2), we have (1) (2) (3) (4) Equation (4) when cleared of fractions, becomes 8x+45-50=66. (5) This gives x=7. Substituting this value of x in (3), we find y=5. 2. Again, suppose we have given, to find x, y, and z, the three equations 2x+4y-3z=22 (1) 4x-2y+5z=18 (2) 6x+7y- z=63. (3) From equation (3) we obtain z=6x+7y-63. (4) Substituting this value of z, in (1) and (2), and they will be come (5) (6) X 2x+4y—3(6x+7-63)=22 4x-2y+5(6x+77–63)=18 Equations (5) and (6) become, after expanding, transposing, and uniting terms, 16x+17y=167 (7) 34x+33y=333. (8) Equation (7) gives 167–174 (9) 16 This value of x, substituted in (8), gives 34(167–174) +33y=333. (10) 16 Equation (10), when solved as a simple equation of one unknown quantity, gives y=7. Substituting this value of y in (9), we find x=3. Using these values of x and y in (4), we obtain (86.) This method of eliminating may be comprehended in the following Having found the value of one of the unknown quantities, from either of the given equations, in terms of the other unknown quantities, substitute it for that unknown quantity in the remaining equations, and we shall thus obtain a new system of equations one less in number than those given. Operate with these new equations as with the first, and so continue until we find one single equation with but one unknown quantity, which will then become known. EXAMPLES. x-w= 50 (1)) 3y- x=120 (2) 1. Given to find w, x, y, and z. 2z-y=120 (3) 3w-z=195 From (1) we find w=x-50. (5) This value of w, substituted in (4), gives 3(x–50)-z=195, or 3.c-z=345. (6) Equation (6) gives z=30-345. (7) This value of z, substituted in (3), gives 2(3x—345)-y=120 or 6x-y=810. (8) Equation (8) gives y=6x—810. This value of y, substituted in (2), gives 3(6.6-810)-x=120 (10) 17x=2550. (11) ..X=150. (9) 4 or This value of x causes (9) to become y=90. Using the value of x in (7), we find z=105. Finally, using the value of x in (5), we find w=100. to find x, y, (3) =a. S x+2y=a 2. Given y+z=a (2) and z+ix=a Equation (3) gives 4a-x (4) 4 This value of z, substituted in (2), we have 40-X yt. (5) 12 Clearing fractions and uniting terms, (5) becomes 12y-x=8a. From (6) we find x=12y-8a. (7) This value of x, substituted in (1), gives Y (8) 2 Equation (8) gives 25y=18a, (9) 18a .'. YS 25" This value of y, substituted in (7), gives 16a Substituting for x, in (4), its value just found, we have 21a |